To tackle the problem step by step:
(a) Gravitational Potential Energy of the Diver
The formula for gravitational potential energy (PE) is given by:
\[ PE = m \cdot g \cdot h \]
where:
- \( m = 65 , \text{kg} \) (mass of the diver)
- \( g = 9.8 , \text{N/kg} \) (acceleration due to gravity)
- \( h = 10.0 , \text{m} \) (height of the tower above water)
Substituting in the values:
\[ PE = 65 , \text{kg} \cdot 9.8 , \text{N/kg} \cdot 10.0 , \text{m} = 6370 , \text{J} \]
So, the gravitational potential energy of the diver relative to the height of the water is 6370 J.
(b) Kinetic Energy of the Diver
The kinetic energy (KE) of the diver as they hit the water can be calculated using the formula:
\[ KE = \frac{1}{2} m v^2 \]
where:
- \( v = 14 , \text{m/s} \)
Substituting in the values:
\[ KE = \frac{1}{2} \cdot 65 , \text{kg} \cdot (14 , \text{m/s})^2 = \frac{1}{2} \cdot 65 \cdot 196 = \frac{1}{2} \cdot 12740 = 6370 , \text{J} \]
So, the kinetic energy of the diver just before entering the water is also 6370 J.
(c) Comparison of Potential and Kinetic Energy
The gravitational potential energy calculated in part (a) and the kinetic energy calculated in part (b) are equal, both being 6370 J. This relationship illustrates the principle of conservation of mechanical energy, which states that in the absence of non-conservative forces (like air resistance), the total mechanical energy of an object remains constant.
In this case, when the diver stands on the tower, they possess gravitational potential energy due to their height. As they dive, this potential energy converts into kinetic energy, leading to a speed of 14 m/s just before hitting the water. The equality of the two energies confirms the conversion of energy and the conservation law.
(d) Speed of the Diver 5.0 m Above the Water
To find the speed of the diver 5.0 m above the water, we can use the conservation of energy once again.
At 5.0 m high, the diver has both potential energy and kinetic energy. The potential energy at that height is:
\[ PE' = m \cdot g \cdot h' = 65 , \text{kg} \cdot 9.8 , \text{N/kg} \cdot 5.0 , \text{m} = 65 \cdot 9.8 \cdot 5 = 3190 , \text{J} \]
The total energy when diving starts at the top (which is also the kinetic energy when about to enter the water) is:
\[ E_{\text{total}} = PE + KE = 6370 , \text{J} + 0 , \text{J} = 6370 , \text{J} \]
At 5.0 m above the water, the kinetic energy can be expressed as:
\[ KE' = E_{\text{total}} - PE' = 6370 , \text{J} - 3190 , \text{J} = 3180 , \text{J} \]
Now, we can solve for the speed at that height using the kinetic energy formula:
\[ KE' = \frac{1}{2} m v'^2 \]
Substituting:
\[ 3180 , \text{J} = \frac{1}{2} \cdot 65 , \text{kg} \cdot v'^2 \]
Solving for \( v' \):
\[ v'^2 = \frac{2 \cdot 3180}{65} \approx \frac{6360}{65} \approx 97.54 \]
\[ v' \approx \sqrt{97.54} \approx 9.87 , \text{m/s} \]
Thus, the speed of the diver 5.0 m above the water is approximately 9.87 m/s.